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Last edited by 青青子衿 2023-2-8 15:28\begin{align*}
\int_{0}^{u}\dfrac{\mathrm{d}t}{\sqrt{\alpha+\beta t+\gamma t^{2}}}+\int_{0}^{v}\dfrac{\mathrm{d}t}{\sqrt{\alpha+\beta t+\gamma t^{2}}}=\int_{0}^{w}\dfrac{\mathrm{d}t}{\sqrt{\alpha+\beta t+\gamma t^{2}}}
\end{align*}
\begin{align*}
w={}&\frac{\beta\left[\left(\sqrt{\alpha+\beta\,\!u+\gamma\,\!u^2}+\sqrt{\alpha+\beta\,\!v+\gamma\,\!v^2}-\sqrt{\alpha }\right)^2-\alpha-\gamma(u-v)^2\right]}{\beta ^2-4 \alpha\gamma}\\
&\qquad-\dfrac{4\gamma\left[u\sqrt{\alpha \left(\alpha+\beta\,\!v+\gamma\,\!v^2\right)}
+v \sqrt{\alpha\left(\alpha+\beta\,\!u+\gamma\,\!u^2\right)}\right]}
{{\beta ^2-4 \alpha\gamma}}
\end{align*}
\begin{align*}
\color{black}{\int_{0}^{u}\dfrac{\mathrm{d}t}{\alpha+\beta t+\gamma t^{2}}+\int_{0}^{v}\dfrac{\mathrm{d}t}{\alpha+\beta t+\gamma t^{2}}=\int_{0}^{w}\dfrac{\mathrm{d}t}{\alpha+\beta t+\gamma t^{2}}}
\end{align*}
\begin{align*}
\color{black}{w=\frac{\alpha u+\alpha v+\beta uv}{\alpha-\gamma uv}}
\end{align*}
\begin{align*}
&\color{black}{ \int_{\frac{13-\sqrt{41}}{8}}^{1}\frac{\mathrm{d}t}{\sqrt{\left(1-t^{2}\right)\left(4-t^{2}\right)}}=\int_{2}^{\frac{13+\sqrt{41}}{8}}\frac{\mathrm{d}t}{\sqrt{\left(1-t^{2}\right)\left(4-t^{2}\right)}} }\\
\\
&\quad\>\>\>\color{black}{\int_{4}^{2+\frac{2\sqrt{14}}{3}\cos\left[\frac{1}{3}\arctan\left(\frac{\sqrt{719}}{45}\right)\right]}\frac{\mathrm{d}t}{\sqrt{(t+3)(t+1)(t-2)(t-4)}}\,}\\
&\color{black}{=\int_{2+\frac{2\sqrt{14}}{3}\cos\left[\frac{2\pi}{3}+\frac{1}{3}\arctan\left(\frac{\sqrt{719}}{45}\right)\right]}^{2+\frac{2\sqrt{14}}{3}\cos\left[\frac{4\pi}{3}+\frac{1}{3}\arctan\left(\frac{\sqrt{719}}{45}\right)\right]}\frac{\mathrm{d}t}{\sqrt{(t+3)(t+1)(t-2)(t-4)}}}
\end{align*} |
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